Proof: Suppose that $K$ is a compact subset of $M.$ To show that $(K,d^{\prime })$ is a compact metric space, let $\{U_{\alpha }|\alpha \in A\}$ be a collection of open subsets in the metric space $(K,d^{\prime })$ that covers $K.$ We need to find a finite subcover of $K$ from $\{U_{\alpha }|\alpha \in A\}.$ By Theorem 2.1, for each $\alpha ,$ there is an open subset ${\mathcal{O}}_{\alpha }$ of $M$ such that $U_{\alpha }=K\cap {\mathcal{O}}_{\alpha }.$ The collection $\{{\mathcal{O}}_{\alpha }|\alpha \in A\}$ is then an open cover of $K$ in $M.$ Since $K$ is a compact subset of $M,$ there is finite subcover $\{{\mathcal{O}}_{{\alpha }_1},{\mathcal{O}}_{{\alpha }_2},\dots ,{\mathcal{O}}_{{\alpha }_n}\}$ of $K$ from $\{{\mathcal{O}}_{\alpha }|\alpha \in A\}.$ But then $\{U_{{\alpha }_1},U_{{\alpha }_2},\dots ,U_{{\alpha }_n}\}$ is a finite subcover of $K$ from $\{U_{\alpha }|\alpha \in A\}.$ To see this, let $k\in K.$ Then there is some $i$ such that $k\in {\mathcal{O}}_{{\alpha }_i}.$ Since $k\in K$ and $k\in {\mathcal{O}}_{{\alpha }_i},$ then $k\in K\cap {\mathcal{O}}_{{\alpha }_i}.$ That is, $k\in U_{{\alpha }_i}.$ So, every element of $k$ is contained in some $U_{{\alpha }_i}.$

Conversely, suppose that $(K,d^{\prime })$ is a compact metric space, and let $\{{\mathcal{O}}_{\alpha }|\alpha \in A\}$ be an open cover of $K$ in $M,$ so that each ${\mathcal{O}}_{\alpha }$ is an open subset of $M.$ Then, again using Theorem 2.1, $\{K\cap {\mathcal{O}}_{\alpha }|\alpha \in A\}$ is an open cover of $K$ consisting of open sets in $(K,d^{\prime }).$ This open cover has a finite subcover $\{K\cap {\mathcal{O}}_{{\alpha }_i}|i=1,2,\dots ,n\}.$ And it is then clear that $\{{\mathcal{O}}_{{\alpha }_i}|i=1,2,\dots ,n\}$ is a finite subcover of $K$ from $\{{\mathcal{O}}_{\alpha }|\alpha \in A\}.$ ∎

Proof: Let $a,b\in \mathbb{R}$ with $a<b,$ and consider the bounded, closed interval $[a,b].$ Suppose that $\mathcal{C}=\{{\mathcal{O}}_{\alpha }|\alpha \in A\}$ is an open cover of $[a,b]$ in $\mathbb{R}.$ Consider the set $X=\{x\in [a,b]|$the interval $[a,x]$ has a finite subcover from $\mathcal{C}\}.$ Now, $a\in {\mathcal{O}}_{\beta }$ for some $\beta ,$ and therefore $\{{\mathcal{O}}_{\beta }\}$ is a finite cover of $[a,a]$ from $\mathcal{C}.$ So, $a\in X,$ and $X$ is non-empty. Furthermore $X$ is bounded above by $b.$ By the least upper bound axiom, every non-empty subset of $\mathbb{R}$ that is bounded above has a least upper bound. Let $\lambda$ be the least upper bound of $X.$ Note that $a\le \lambda \le b$ because $a\in X$ and $b$ is an upper bound for $X.$ That is, $\lambda \in [a,b].$

We show that $\lambda \in X$ and that $\lambda =b.$ This will finish the proof of the theorem, since by definition of $X,$ it will mean that $[a,b]$ has a finite subcover from $\mathcal{C}.$

Since $\lambda \in [a,b],$ and $\mathcal{C}$ is an open cover of $[a,b],$ there is a $\beta \in A$ such that $\lambda \in {\mathcal{O}}_{\beta }.$ Since ${\mathcal{O}}_{\beta }$ is open, there is an $\epsilon >0$ such that $(\lambda -\epsilon ,\lambda +\epsilon )\subseteq {\mathcal{O}}_{\beta }.$ Since $\lambda$ is the least upper bound for $X,$ there must be some $x\in X$ such that $x>\lambda -\epsilon ,$ for otherwise, $\lambda -\epsilon$ would be an upper bound for $X$ that is smaller than $\lambda .$ Since $x\in X,$ there is a finite subcover of $[a,x]$ from $\mathcal{C},$ say $\{{\mathcal{O}}_{{\alpha }_1},{\mathcal{O}}_{{\alpha }_2},\dots ,{\mathcal{O}}_{{\alpha }_n}\}.$ But then since $[x,\lambda ]\subseteq {\mathcal{O}}_{\beta },$ we get a finite subcover of $[a,\lambda ]$ by adding ${\mathcal{O}}_{\beta }$ to $\{{\mathcal{O}}_{{\alpha }_1},{\mathcal{O}}_{{\alpha }_2},\dots ,{\mathcal{O}}_{{\alpha }_n}\}.$ So, by definition of $X,$ $\lambda \in X.$

Finally, we show that $\lambda =b.$ Suppose for the sake of contradiction that $\lambda <b.$ Let $y=\lambda +\frac{1}{2}min(\epsilon ,b-\lambda ).$ Then $y\in {\mathcal{O}}_{\beta }$ and therefore $\{{\mathcal{O}}_{\beta },{\mathcal{O}}_{{\alpha }_1},{\mathcal{O}}_{{\alpha }_2},\dots ,{\mathcal{O}}_{{\alpha }_n}\}$ is a finite subcover of $[a,y]$ from $\mathcal{C}.$ And $y\in [a,b],$ because $\lambda <y<b.$ This means $y\in X$ and $\lambda <y.$ But that contradicts the fact that $\lambda$ is an upper bound for $X.$ This contradiction shows that $\lambda =b,$ which finishes the proof. ∎

Proof: We show $K$ is closed by showing that its complement, $M\setminus K,$ is open. Let $z\in M\setminus K.$ We must find an $\epsilon >0$ such that $B_{\epsilon }(z)\subseteq M\setminus K.$ For each $x\in K,$ let ${\epsilon }_x=\frac{1}{2}d(x,z).$ Since $z\notin K,$ ${\epsilon }_x>0,$ so that $B_{{\epsilon }_x}(x)$ is an open set. Note that $x\in B_{{\epsilon }_x}(x),$ and $z\notin B_{{\epsilon }_x}(x).$ The collection $\{B_{{\epsilon }_x}(x)|x\in K\}$ is an open cover of $K,$ since any $x\in K$ is covered by the open set $B_{{\epsilon }_x}(x).$ Since $K$ is compact, there is a finite subcover $\{B_{{\epsilon }_{x_i}}(x_i)|i=1,2,\dots ,n\}.$ Let $\epsilon =min({\epsilon }_{x_1},{\epsilon }_{x_2},\dots ,{\epsilon }_{x_n}\}.$

We show that $B_{\epsilon }(z)\subseteq M\setminus K,$ which will finish the proof. To show this, let $y\in B_{ϵ}(z).$ We want to show $y\in M\setminus K,$ that is, $y\notin K.$ Consider $x_i,$ where $1\le i\le n.$ By the triangle inequality, $d(x_i,y)+d(y,z)\ge d(x_i,z)=2{\epsilon }_{x_i}.$ So, subtracting $d(y,z)$ from both sides of the inequality, $d(x_i,y)\ge 2{\epsilon }_{x_i}-d(y,z)>2{\epsilon }_{x_i}-{\epsilon }_{x_i}={\epsilon }_{x_i}.$ (The last inequality follows because $d(y,z)<\epsilon \le {\epsilon }_{x_i}.$) Then, since $d(x_i,y)>{\epsilon }_{x_i},$ $y$ is not in the open ball of radius ${\epsilon }_{x_i}$ about $x_i.$ Since this is true for every $i$ and the open balls $B_{{\epsilon }_{x_i}}(x_i)$ cover $K,$ we have that $y\notin K.$ ∎

Proof: If $K$ is empty, then it is bounded. Suppose that $K$ is not empty. Let $x$ be any element of $K,$ and consider the collection of open balls of integral radius, $\{B_i(x)|i=1,2,\dots \}.$ Since every element of $K$ has some finite distance from $x,$ this collection is an open cover of $K,$ Since $K$ is compact, it has a finite subcover $\{B_{i_1}(x),B_{i_2}(x),\dots ,B_{i_n}(x)\},$ where we can assume $i_1<i_2<\cdots <i_n.$ But since $B_{i_1}(x)\subseteq B_{i_2}(x)\subseteq \cdots \subseteq B_{i_n}(x),$ this means that $B_{i_n}$ by itself already covers $K.$ Then, for $y,z\in K,$ $y$ and $z$ are in $B_{i_n}(x),$ and $d(y,z)\le d(y,x)+d(x,z)\le i_n+i_n.$ It follows that $2i_n$ is an upper bound for $\{d(y,z)|y,z\in K\}.$ So, $K$ is bounded. ∎

Proof: We prove the contrapositive. Let $X\subseteq K.$ Assume that $X$ has no accumulation point in $K.$ We must show that $X$ is finite. Let $z\in K.$ Since $z$ is not an accumulation point of $X,$ there is an ${\epsilon }_z>0$ such that $X\cap (B_{{\epsilon }_z}(z)\setminus \{z\})=\varnothing .$ Thus, $X\cap B_{{\epsilon }_z}(z)$ either is empty or is the single point $\{z\}.$ The set of open balls $\{B_{{\epsilon }_z}(z)|z\in K\}$ is an open cover of $K.$ Since $K$ is compact, there is a finite subcover. That is, there are finitely many points $z_1,z_2,\dots ,z_n$ such that the corresponding open balls already cover $K.$ But each of the $n$ open balls in that subcover contains at most one point of $X,$ and it follows that $X$ has $n$ or fewer points. So we have proved that $X$ is finite. ∎

Proof: If the terms of the sequence include only finitely many different points of $K,$ then at least one of those points, say $x,$ occurs infinitely often in the sequence. Then $\{x_i{\}}_{i=1}^{\mathrm{\infty }}$ has a constant subsequence consisting of all the occurrences of $x,$ and that subsequence converges to $x,$ which is in $K$ since every $x_i$ is in $K.$

So, suppose that the terms of the sequence include infinitely many different points in $K.$ Since $K$ is compact, the infinite subset containing all of the terms of the sequence has an accumulation point in $K,$ by the previous theorem. Let $z$ be that accumulation point. We show that $\{x_i{\}}_{i=1}^{\mathrm{\infty }}$ has a subsequence that converges to $z.$ We use the result from Exercise 1.8, which implies that for any $ϵ>0,$ the open ball $B_{ϵ}(z)$ contains infinitely many of the terms of the sequence. Since $z$ is an accumulation point of the set consisting of all the $x_i,$ there is an index $i_1$ such that $x_{i_1}\in B_1(z).$ Then, since $B_{1/2}(z)$ contains infinitely many terms of the sequence, there is an index $i_2>i_1$ such that $x_{i_2}\in B_{1/2}(z).$ Then, since $B_{1/3}(z)$ contains infinitely many terms of the sequence, there is an index $i_3>i_2$ such that $x_{i_3}\in B_{1/3}(z).$ Proceeding in this way, we obtain indices $i_1<i_2<i_3<\cdots$ such that $x_{i_n}\in B_{1/n}(z)$ for all $n\in \mathbb{N}.$ It follows easily that the subsequence $\{x_{i_n}{\}}_{n=1}^{\mathrm{\infty }}$ converges to $z.$ ∎

Proof: Let $K$ be a compact subset of $A.$ To show that $f(K)$ is compact, we must show that any open cover of $f(K)$ has a finite subcover. Suppose that $\{{\mathcal{O}}_{\alpha }|\alpha \in I\}$ is an open cover of $f(K)$ in $B.$ By Theorem 3.3, each set $f^{-1}({\mathcal{O}}_{\alpha })$ is open in $A.$ Furthermore, the collection $\{f^{-1}({\mathcal{O}}_{\alpha })|\alpha \in I\}$ covers $K$, since for $k\in K,$ $f(k)\in {\mathcal{O}}_{\beta }$ for some $\beta ,$ and then $k\in f^{-1}({\mathcal{O}}_{\beta }).$ Since $K$ is compact, it has a finite subcover, $\{f^{-1}({\mathcal{O}}_{{\alpha }_1}),f^{-1}({\mathcal{O}}_{{\alpha }_2}),\dots ,f^{-1}({\mathcal{O}}_{{\alpha }_n})\}.$ I claim that $\{{\mathcal{O}}_{{\alpha }_1},{\mathcal{O}}_{{\alpha }_2},\dots ,{\mathcal{O}}_{{\alpha }_n}\}$ is a finite subcover of $f(K)$ from $\{{\mathcal{O}}_{\alpha }|\alpha \in I\}.$ In fact, let $b\in f(K).$ That is, $b=f(a)$ for some $a\in K.$ Now, $a\in f^{-1}({\mathcal{O}}_{{\alpha }_i})$ for some $i.$ So, $b=f(a)\in f(f^{-1}({\mathcal{O}}_{{\alpha }_i}))\subseteq {\mathcal{O}}_{{\alpha }_i}.$ That is, every element of $f(K)$ is covered by some ${\mathcal{O}}_{{\alpha }_i}.$ ∎